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Nombre de lectures 42
Langue English

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Physics Summary

Contents Page

Core Topic One: Space

1. Gravity 2
2. Space Launch and Return 3
3. Future Space Travel 13
4. Special Relativity 14

Core Topic Two: Motors and Generators

1. The Motor Effect 19
2. Electromagnetic Induction 24
3. Electric Generators 27
4. Transformers 29
5. Electric Motors 31

Core Topic Three: From Ideas to Implementation

1. Cathode Rays 32
2. Quantum Theory 37
3. Solid State Devices 43
4. Superconductivity 48

Option Topic: Quanta to Quarks

1. Models of the Atom 53
2. Quantum Physics 57
3. The Electron Microscope 59
4. Applications of Radioactivity 61
5. Nuclear Applications 66
6. The Structure of Matter 67

William Kim HSC Physics Summary | page 1 ç
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Core Topic One: Space

1. The Earth has a gravitational field that exerts a force on objects both on it and around it

§ Define weight as the force on The weight of an object is the force of gravity acting on it.
r ran object due to a W = mg
gravitational field
Where W is the weight in newtons (N), m is the mass in kilograms (kg) and g can be either:
1. The acceleration due to gravity (= 9.8 m/s/s at the Earth’s surface); or
2. The gravitational field strength (= 9.8 N/kg at the Earth’s surface).


§ Define gravitational potential As we lift an object from the ground to a height above the ground we do work on it. This work is
energy as the work done to stored in the object as gravitational potential energy. For an object of mass m at a height h above
move an object from a very the Earth’s surface the gravitational potential energy E is given by:
large distance away to a point However this equation is valid only when the object is near the Earth’s surface. E = mghp
in a gravitational field.
The gravitational potential energy is a measure of the work done in moving an object from
infinity to a point in the field. The general expression for the gravitational potential energy of an
object of mass m at a distance r from the centre of the Earth (or other planet) is given by: Newton’s Law of Universal
Gravitation mM EE = -G Where M is the mass of the Earth (or other planet). p rm m1 2 F = G
2r Change in Gravitational Potential Energy
The change in potential energy of a mass m as it moves from infinity to a distance r from a source 1where G is the universal
of a gravitational field (due to a mass m ) is given by: 2gravitational constant.
m m 1 2DE = G pThe Gravitational Field r
Surrounding any object with
mass is a gravitational field. Change in Gravitational Potential Energy Near the Earth (when radius increases from A to B)

æ 1 1Gm DE = GmM -g = p E2 r rr Ł A B ł
William Kim HSC Physics Summary | page 2 ç
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2. Many factors have to be taken into account to achieve a successful rocket launch, maintain a stable orbit and return to Earth

§ Describe the trajectory of an Any moving object that moves only under the force of gravity is a projectile. The horizontal motion
object undergoing projectile of a projectile is independent to the vertical motion. The reason for this result is that gravity is the
motion within the Earth’s only force acting on the objects and this always acts towards the centre of the Earth.
gravitational field in terms of
horizontal and vertical Projectile motion can be analysed by realising that:
components 1. The horizontal motion is constant velocity.
2. The vertical motion of constant acceleration (with acceleration of g).

u Equations of Uniformly Accelerated Motion x

r r r
v = u + at
Dy
r r 1 r 2s = ut + at
2
Dx 2 2v = u + 2as

The Path of a Projectile
The velocity at any point of the path of a projectile is simply the vector sum of the horizontal and
vertical velocity components at that point.

2 The horizontal component is constant. Dy=k(Dx)
The vertical component changes at g, the acceleration due to gravity.
aæ g k=
22u Trajectories Ł xł
The path followed by a projectile – its trajectory – is a parabola (or linear)

From (1): Combining (2) & (3): (1) Horizontal motion: Dx = u t
x
2Dx a1 æ1 Dx 12 g 2t = (2) Vertical motion: Dy = a t Dy= a = ()Dx g g 2u2 x 2 u 2uŁ xł x
William Kim HSC Physics Summary | page 3
§ Describe Galileo’s analysis of Galileo was responsible for deducing the parabolic shape of the trajectory of a projectile. Galileo’s
projectile motion analysis of projectile motion led him to consider reference frames. These are what all measurements
are compared to.
The concept of Galilean relativity refers that the laws of mechanics are the same in a frame
of reference that is at rest or one that moves with constant velocity.


§ Explain the concept of escape If an object is projected upward with a large enough velocity it can escape the gravitational pull of
velocity in terms of the: the Earth (or other planet) and go into space. The necessary velocity to leave the Earth (or other
planet) is called the escape velocity.
o gravitational constant Escape velocity depends on the gravitational constant, the mass and radius of the planet.
o mass and radius of the
planet Suppose an object of mass m is projected vertically upward from the Earth’s surface (mass of M and
radius R) with an initial velocity u. The initial mechanical energy, that is, kinetic and potential
energy is given by:
1 M m2 EE + E = mu - Gk pi i 2 RE

Let us assume that the initial speed is just enough so that the object reaches infinity with zero
velocity. The value of the initial velocity for which this occurs is the escape velocity v . e

When the object is at infinity the mechanical energy is zero (the kinetic energy is zero since the
velocity is zero and the potential energy is zero because this is where we selected the zero of
potential energy).

1 M m2 EHence which leads to: mv - G = 0e2 RE
2GM
Ev =e
RE
William Kim HSC Physics Summary | page 4
×
§ Discuss Newton’s analysis of Circular Motion
escape velocity The motion of an object in a circular path with constant speed is called uniform circular motion.
Although the speed remains the same in uniform circular motion, it follows that an object travelling
in a circular path must be accelerating, since the velocity (that is, the speed in a given direction) is
continually changing. r r
v „ v v = v
1 2 1 2
r
v2
r The change in velocity is given by: vr 2 r r r v Dv = v - v1 2 1
r v1 r r Dvr a =and since: Dv r DtDv

it follows that the object is accelerating.
Isaac Newton proposed the idea
of artificial satellites of the

Earth. He considered how a Centripetal Acceleration
projectile could be launched As can be seen, when the change in velocity is placed in the average position between v and 1horizontally from the top of a v , it is directed towards the centre of the circle. When an object is moving with uniform circular 2
high mountain so that it would
motion, the acceleration (the centripetal acceleration) is directed towards the centre of the circle.
not fall to Earth. For an object moving in a circle of radius r with an orbital velocity of v, the centripetal acceleration
As the launch velocity a is given by: 2was increased, the distance that v a =cthe object would travel before r
hitting the Earth would increase Earth Orbits
until such a time that the
A satellite can be put into Earth orbit by lifting it to a sufficient height and then giving it the
velocity would be sufficient to required horizontal velocity so that it does not fall back to Earth. For the satellite to circle the Earth,
put the object into orbit around the centripetal force required is provided by the gravitational attraction between the satellite and the
the Earth. (A higher velocity
Earth. Hence the centripetal acceleration is given by: 2would lead to the object v
g =escaping from the Earth.) R
William Kim HSC Physics Summary | page 5
§ Use the term ‘g forces’ to The human body is relatively unaffected by high speeds. Changes in speed, however, that is,
explain the forces acting on accelerations, can and do affect the human body creating ‘acceleration stress’.
an astronaut during launch
g-forces
Acceleration forces – g-forces – are measured in units of gravitational acceleration g. For example, g-forces on Astronauts
a force of 5g is equivalent to acceleration five times the acceleration due to gravity. Humans can withstand 4g
without undue concern.
If the accelerations are along the body’s long axis then two distinct effects are possible: Accelerations up to ~10g are
1. If the acceleration is in the direction of the person’s head they may experience a ‘black out’ tolerable for short times when
as the blood rushes to their feet; or the acceleration is directed
2. If the acceleration is towards their feet, they may experience a ‘red out’ where the blood parallel to a line drawn between
rushes to their head and retina. the person’s front and back.


As you ‘fall’ from a height, you experience negative g-forces (you feel lighter). When you ‘pull out’ § Compare the forces acting on
of a dip after a hill or follow an ‘inside loop’, you experience positive g-forces (you feel heavier). an astronaut during launch
The positive g-forces are like those astronauts experience at lift-off. with what happens during a
roller coaster ride
Consider a rider in a car at the bottom of an inside loop. The rider has two forces acting on them:
1. Their normal weight (mg) acting down; and
2. The ‘normal reaction force’ (N) acting up. This is the push of the seat upwards on their
N bottom.
Assume that the loop is part of a circle of radius R. A centripetal force is required for the rider to
travel in a circle. This is the difference between the normal force and the weight force, that is:
22mv mv
N - mg = : N = mg +
mg R R
The g-forces are found from the 'normal force’ divided by the weight. That is:

2mv mg + 2N vRg’s felt by rider = = = 1+
mg mg gR
William Kim HSC Physics Summary | page 6
§ Discuss the impact of the A moving platform offers a boost to the velocity of a projectile launched from it, if launched in the
Earth’s orbital motion and its direction of motion of the platform. This principle is used in the launch of a rocket by considering
rotational motion on the that the Earth revolves around the Sun at 107,000km/h relative to the Sun and rotates once on its
launch of a rocket axis per day so that a point on the Equator has a rotational velocity of approximately 1,700km/h
relative to the Sun. Hence, the Earth is itself a moving platform with two different motions which
can be exploited in a rocket launch to gain a boost in velocity.

Earth Orbit
A rocket heading into orbit is launched to the east to receive a velocity boost from the Earth’s
rotational motion.

An Interplanetary Trip
The flight of a rocket heading into space is timed so that it can head out in the direction of the
Earth’s motion and thereby receive an extra boost.


§ Analyse the changing Law of Conservation of Momentum
acceleration of a rocket Rocket engines generate thrust by burning fuel and expelling the resulting gases. Conservation of
during launch in terms of the: momentum means that as the gases move one way, the rocket moves the other. (Momentum before
the burning is zero; hence the momentum after is also zero. The gases carry momentum in one
Law of Conservation of direction down, and so the rocket carries an equal momentum in the opposite direction up.)
Momentum As fuel is consumed and the gases expelled, the mass of the system decreases. Since
acceleration is proportional to the thrust and inversely proportional to the mass, as the mass
Forces experienced by decreases, the acceleration increases. Hence the forces on the astronauts increase.
astronauts
Forces Experienced by Astronauts
g forces varied during the launch of Saturn V, a large three-stage rocket used to launch the Apollo
spacecraft. This is attributed to the sequential shutdown of the multiple rocket engines of each stage
– a technique designed specifically to avoid extreme g forces.
William Kim HSC Physics Summary | page 7
§ Analyse the forces involved Motion Fc Provided By…
in uniform circular motion Whirling rock on a string The string
for a range of objects,
Electron orbiting atomic nucleus Electron-nucleus electrical attraction including satellites orbiting
the Earth Car cornering Friction between tyres and road

Moon revolving around Earth Moon-Earth gravitational attraction
Satellite revolving around Earth Satellite-Earth gravitational attraction



§ Compare qualitatively and Low Earth Orbit
quantitatively low Earth and A low Earth orbit is generally an orbit higher than approximately 250 km, in order to avoid
geostationary orbits atmospheric drag, and lower than approximately 1000 km, which is the altitude at which the Van
Allen radiation belts start to appear. The space shuttle utilises a low Earth orbit somewhere between
250 km and 400 km depending upon the mission. At 250 km, an orbiting spacecraft has a velocity
of 27,900km/h and takes just 90 minutes to complete an orbit of the Earth.

Geostationary Orbit
A geostationary orbit is at an altitude at which the period of the orbit precisely matches that of the
Earth. If over the Equator, such an orbit would allow a satellite to remain ‘parked’ over a fixed
point on the surface of the Earth throughout the day and night. From the Earth such a satellite
appears to be stationary in the sky, always located in the same direction regardless of the time of
day. This is particularly useful for communications satellites because a receiving dish need only
point to a fixed spot In the sky in order to remain in contact with the satellite.
The altitude of such an orbit is approximately 38,800 km. If a satellite at this height is not
positioned over the Equator but at some other latitude, it will not remain fixed at one point in the
sky. Instead, from the Earth the satellite will appear to trace out a ‘figure of eight’ path each 24
hours. It still has a period equal to the Earth’s, however, and so this orbit is referred to as
geosynchronous.
William Kim HSC Physics Summary | page 8
§ Discuss the important of Once a launched rocket has achieved a sufficient altitude above the surface of the Earth, it can be
Newton’s Law of Universal accelerated into an orbit. It must attain a specific speed that is dependent only upon the mass and
Gravitation in understanding radius of the Earth and the altitude above it. If that speed is not reached, the spacecraft will spiral
and calculating the motion of back in until it re-enters the atmosphere; if the speed is exceeded, it will spiral out. This can be
satellites considered by appreciating that the simplest orbital motion is a uniform speed along a circular path
around the Earth.
Uniform circular motion, as already mentioned, is a circular motion with a uniform orbital
Using Newton’s Law of velocity. According to Newton’s First Law of Motion, a spacecraft in orbit around the Earth, or any
Universal Gravitation combined object in circular motion, requires some force to keep it there, otherwise it would fly off at a tangent
with the expression for to the circle. This force is directed back towards the centre of the circle. In the case of spacecraft, it
centripetal force, we can see is the gravitational attraction between the Earth and the spacecraft that acts to maintain the circular
that the orbital velocity motion that is the orbit. The force required to maintain circular motion, known as centripetal force,
2required for a particular orbit can be determined using the following equation: mv
F =depends only on the mass of the C rEarth, the radius of the Earth
and the altitude of the orbit The application of Newton’s Law of Universal Gravitation to the orbital motion of a
(distance from the surface of the satellite will produce an expression for the critical orbital velocity mentioned earlier. Recall that this
Earth). Given that the mass and law states that the gravitational attraction between a satellite and the Earth would be given by the
radius of the Earth have fixed following expression: m mE Svalues, this means that altitude F = GG 2ris the only variable that
determines the specific velocity This gravitational force of attraction also serves as the centripetal force for the circular orbital
required. In addition, the motion, hence: F = FG Cgreater the radius of the orbit,
the lower the orbital velocity Therefore, we can equate the formula for F with that for F : G C
required.
2m m m v where E S S
-1G =
2 v = orbital velocity (ms ) r r
Gm where E\v = r = r + altitude (m)E r
William Kim HSC Physics Summary | page 9
Further, we can use the expression for orbital velocity to prove Kepler’s Third Law – the Law of Kepler’s Third Law:
Periods. The period or the time taken to complete one full orbit can be found by dividing the length The Law of Periods
of the orbit (the circumference of the circle) by the orbital velocity, v.
2pr
T =
v

Changing the subject of this expression to v and then substituting into the formula for v given
above:
2pr Gm E= T r
3 Gmr E \ =
2 2T 4p
3 2This means that for any satellite of the Earth at any altitude, the ratio r :T always equals the same
fixed value.
William Kim HSC Physics Summary | page 10